Problem 47 - Algebra and Trigonometry

Question

a. Use the Leading Coefficient Test to determine the graph's end behavior. b. Find the \(x\)-intercepts. State whether the graph crosses the \(x\)-axis, or touches the \(x\) -axis and turns around, at each intercept. c. Find the \(y\)-intercept. d. Determine whether the graph has y-axis symmetry, origin symmetry, or neither. e. If necessary, find a few additional points and graph the function. Use the maximum number of turning points to check whether it is drawn correctly. \(f(x)=x^{4}-2 x^{3}+x^{2}\)

Step-by-Step Solution

Answer

The end behavior of the graph is \(\infty\) as \(x \rightarrow \pm \infty\). The x-intercepts are at x=0 and x=1 where the graph crosses the x-axis. The y-intercept is at y=0. The function doesn't have y-axis symmetry nor origin symmetry. This information helps in accurately graphing the function.

Step 1: Apply Leading Coefficient Test

Determine the leading coefficient (the coefficient of the term with the highest power of \(x\)) and its degree. The leading coefficient here is 1 and degree of the polynomial is 4. Since the degree is even and the leading coefficient is positive, as \(x\) goes to \(\pm \infty\), \(f(x)\) also goes to \(+\infty\) (following Leading Coefficient Test). This indicates that the graph rises to the far left and far right.

Step 2: Find the x-intercepts

The function \(f(x)\) crosses the \(x\)-axis where f(x) = 0. To find the x-intercepts, set \(f(x) = x^{4}-2 x^{3}+x^{2} = 0\). We can factor out an \(x^{2}\) to get \(x^{2}(x^{2} - 2x + 1) = 0\). Solving this gives x=0 and x=1. Hence there are two x-intercepts - at x=0 and x=1. The graph crosses the x-axis at these points because the multiplicities of the zeros are not even.

Step 3: Find the y-intercept

The y-intercept of the function is found by setting \(x=0\) in the function. So the y-intercept is at \(f(0) = (0)^{4}-2(0)^{3}+(0)^{2} = 0\). This means the function intersects y-axis at y = 0.

Step 4: Check for Symmetry

The function is symmetric with respect to the y-axis if it is an even function and symmetric with respect to the origin if it is an odd function. This function is not symmetric to the y-axis nor the origin because it is not an even or odd function. Therefore, this function has neither y-axis symmetry nor origin symmetry.

Step 5: Graph the function

Graph the function to confirm the results. You can select a few additional points if needed for graphing the function.

Key Concepts

Leading Coefficient Test

The Leading Coefficient Test is a handy tool that helps us determine the end behavior of polynomial functions. It focuses on two components: the leading coefficient, which is the coefficient of the term with the highest power of \(x\), and the degree of the polynomial. In the provided function \(f(x)=x^{4}-2x^{3}+x^{2}\), the leading coefficient is 1, and the degree is 4. This degree is even, and the leading coefficient is positive.

When the degree is even and the leading coefficient is positive, both ends of the graph will rise. That means as \(x\) approaches both positive and negative infinity, \(f(x)\) approaches positive infinity. In simpler terms, the graph will start low on the left, rise, and continue rising on the right. Understanding this pattern is crucial for predicting how the graph will look at its extremes.

X-Intercepts

X-intercepts occur where the graph crosses the \(x\)-axis, or in other words, where \(f(x)=0\). For the function \(f(x)=x^{4}-2x^{3}+x^{2}\), we need to set this equation to zero and solve for \(x\). By factoring out \(x^{2}\), we can rewrite it as \(x^{2}(x^{2} - 2x + 1) = 0\).

Solving this gives \(x=0\) and \(x=1\), which are the \(x\)-intercepts. The graph will cross at these points as the multiplicities of the zeros are not even. When multiplicities are odd, the graph crosses the \(x\)-axis; when even, it touches and turns around on the \(x\)-axis. For this function, both intercepts allow the graph to cross the axis.

Y-Intercepts

The \(y\)-intercept of a function is the point where the graph crosses the \(y\)-axis. To find it, we set \(x=0\) in the function. Applying this to \(f(x)=x^{4}-2x^{3}+x^{2}\), we substitute \(x=0\) and compute \(f(0) = (0)^{4} - 2(0)^{3} + (0)^{2} = 0\).

Thus, the \(y\)-intercept is the point where the graph touches the \(y\)-axis at \((0, 0)\). The \(y\)-intercept directly relates to understanding the function's behavior at the center of its graph, a key feature when sketching the overall graph.

Graph Symmetry

Symmetry in graphing helps identify predictable repeated patterns, simplifying the graphing process. A function can show symmetry about the \(y\)-axis (even functions) or the origin (odd functions). To test for symmetry, a common method is substituting \(-x\) for \(x\) and comparing the result to \(f(x)\).

If \(f(-x) = f(x)\), the function is even, which signifies \(y\)-axis symmetry. If \(f(-x) = -f(x)\), the function is odd, indicating origin symmetry. However, for \(f(x)=x^{4}-2x^{3}+x^{2}\), neither condition is satisfied; it is neither symmetric over the \(y\)-axis nor the origin. Recognizing symmetry types can streamline graphing and analysis, but not all functions will possess these symmetries.